Let $f(x)=\cos(x)x^{-2}$. $f'(x)=$
Solution: $f(x)$ is the product of two, more basic, expressions: $\cos(x)$ and $x^{-2}$. Therefore, the derivative of $f$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}(\cos(x)x^{-2}) \\\\ &=\dfrac{d}{dx}(\cos(x))x^{-2}+\cos(x)\dfrac{d}{dx}(x^{-2})&&\gray{\text{The product rule}} \\\\ &=-\sin(x)x^{-2}+\cos(x)(-2)x^{-3}&&\gray{\text{Differentiate }\cos(x)\text{ and }x^{-2}} \\\\ &=-\sin(x)x^{-2}-2\cos(x)x^{-3}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $f'(x)=-\sin(x)x^{-2}-2\cos(x)x^{-3}$ or any other equivalent form.